Asked by Terrence Eko

A 75 kg box slides down a 25.0 degree ramp with an acceleration of 3.60 m/s^2.

Find coefficient between the box and the ramp.

What acceleration would a 175 kg box have on this ramp?
Answered by drwls
The net force down the ramp is
(mass) x (acceleration) = 75*3.60 = 270 N

The component of weight down the ramp is
M g sin 25 = 310.6 M

The friction force is the difference between those two numbers, 40.6 N

The kinetic friction coefficient is
(friction force)/(normal weight component)
= 40.6/[735 cos 25] = 0.06
Answered by Anonymous
1.54
Answered by Anonymous
a)F=ma Fg=mg 310.9-270 cos(25)735
75(3.60) 75(9.81) 41N 667N
735
F parallel= sin(25)735=310.9N
Answered by jamal
George Washington
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